Euler's product formula in number theory

Question

Is there intuitive proof of Euler's product formula in number theory (not searching for probabilistic proof) which is used to compute Euler's totient function?

Answer 1

One intuition uses Inclusion-Exclusion. For example, if $n$ is divisible by three distinct primes $p$, $q$, and $r$, then

$$n-\left({n\over p}+{n\over q}+{n\over r}\right)+\left({n\over pq}+{n\over pr}+{n\over qr} \right)-{n\over pqr}$$

counts all the numbers less than $n$ that are not divisible by $p$, $q$, or $r$, which is to say the numbers that are relatively prime to $n$. But we see that this is equal to the product

$$n\left(1-{1\over p} \right)\left(1-{1\over q} \right)\left(1-{1\over r} \right)$$

Answer 2

Intuitively $$\frac1{1-p_i^{-s}} = 1+p_i^{-s} + \left(p_i^{-s}\right)^2 + \cdots =\left(p_i^0\right)^{-s}+\left(p_i^1\right)^{-s} + \left(p_i^2\right)^{-s} + \cdots$$

so

$$\prod\limits_{i=1}^\infty \frac1{1-p_i^{-s}} \\=\left(\left(2^0\right)^{-s}+\left(2^1\right)^{-s} + \left(2^2\right)^{-s} + \cdots \right) \times \left(\left(3^0\right)^{-s}+\left(3^1\right)^{-s} + \left(3^2\right)^{-s} + \cdots \right) \\ \times \left(\left(5^0\right)^{-s}+\left(5^1\right)^{-s} + \left(5^2\right)^{-s} + \cdots \right) \times \cdots \\=\left(2^0\right)^{-s} \left(3^0\right)^{-s} \left(5^0\right)^{-s}\cdots + \left(2^1\right)^{-s} \left(3^0\right)^{-s} \left(5^0\right)^{-s}\cdots + \left(2^0\right)^{-s} \left(3^1\right)^{-s} \left(5^0\right)^{-s}\cdots \\ + \left(2^2\right)^{-s} \left(3^0\right)^{-s} \left(5^0\right)^{-s}\cdots + \left(2^0\right)^{-s} \left(3^0\right)^{-s} \left(5^1\right)^{-s}\cdots + \cdots \\ = \left(2^0 3^0 5^0\right)^{-s} +\left(2^1 3^0 5^0\right)^{-s} +\left(2^0 3^1 5^0\right)^{-s} +\left(2^2 3^0 5^0\right)^{-s} +\left(2^0 3^0 5^1\right)^{-s} +\cdots \\ = 1^{-s} + 2^{-s} + 3^{-s} +4^{-s} +5^{-s} + \cdots \\ = \sum\limits_{n=1}^\infty \frac{1}{n^s}$$

though for a proof you will want to use convergence and the fundamental theorem of arithmetic

Answer 3

Presumably you mean $\phi(n)=n\prod_{p|n}(1-1/p)$. Well, I think the best intuition is to understand it.

First, let's look at $\phi(p^n)$. It's easy to see that you can multiply $p$ with any number less than $p^{n-1}$, to get precisely all the numbers less than $p^n$ not relatively prime to $p^n$. Thus we get $\phi(p^n)=p^n-p^{n-1}=p^n(1-1/p)$.

From here all we need is the content of a very famous theorem, called the Chinese Remainder Theorem. It implies that the totient function is multiplicative. That is, when $(m,n)=1$, we have $\phi(mn)=\phi(m)\phi(n)$. As to CRT, it is not that difficult given Bezout's identity.

From these two facts the result follows.